## University Calculus: Early Transcendentals (3rd Edition)

The behaviors of the curves in the plotted graph show that the derivative of $f(x)=\sqrt x$ is $\frac{1}{2\sqrt x}$.
The graphs of $y=\frac{1}{2\sqrt x}$ and $y=\frac{\sqrt{x+h}-\sqrt x}{h}$ are plotted and denoted below. We see that as $h$ gets smaller and smaller and approaches $0$, the curve of $y=\frac{\sqrt{x+h}-\sqrt x}{h}$ gets closer and in fact, becomes the same as the curve of $y=\frac{1}{2\sqrt x}$. We can thus deduce that on $[0,2]$: $$\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt x}{h}=\frac{1}{2\sqrt x}$$ According to the definition of derivative, $\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt x}{h}=f'(x)$ for $f(x)=\sqrt x$. Therefore, the behaviors of the curves in the plotted graph show that the derivative of $f(x)=\sqrt x$ is $\frac{1}{2\sqrt x}$.